If TP and TQ are two tangents to a circle with centre O from an external point T so that $\angle POQ = 120°$ , then ∠ PTQ is equal to :

(a) 60° (b) 70° (c) 80° (d) 90°

Q: If TP and TQ are two tangents to a circle with centre O from an external point T so that $\angle POQ = 120°$, then ∠ PTQ is equal to : (a) 60° (b) 70° (c) 80° (d) 90°

(a) 60° — since $\angle OPT = \angle OQT = 90°$, the quadrilateral angle sum gives $90° + 90° + 120° + \angle PTQ = 360°$, so $\angle PTQ = 60°$.

Q: What is the relationship between ∠PTQ and ∠POQ for two tangents from an external point?

$\angle PTQ + \angle POQ = 180°$, because OPTQ is a quadrilateral with two right angles at P and Q (where the radius meets the tangent).

Circles

Circles - Tangents from an External Point

Options

(A)60°

(B)70°

(D)90°

Official Answer

(a) 60° — since $\angle OPT = \angle OQT = 90°$ , the quadrilateral angle sum gives $90° + 90° + 120° + \angle PTQ = 360°$ , so $\angle PTQ = 60°$ .

tangent to a circleexternal pointangle between tangentsradius perpendicular to tangentquadrilateral angle sum60 degrees

Marking Scheme

11 mark: correct option (a) 60°; full credit reasoning: $\angle OPT = \angle OQT = 90°$ , so $90° + 90° + 120° + \angle PTQ = 360° \implies \angle PTQ = 60°$ .

Hint

Use $OP \perp TP$ , $OQ \perp TQ$ , and the angle sum of quadrilateral OPTQ = 360°, or directly apply $\angle PTQ + \angle POQ = 180°$ .

Quick Oral Answer

Since tangent is perpendicular to radius, $\angle OPT = \angle OQT = 90°$ ; using the angle sum of quadrilateral OPTQ = 360°, we get $\angle PTQ = 360° - 90° - 90° - 120° = 60°$ .

Analysis & Explanation

Tests the tangent–radius perpendicularity property together with the angle sum of a quadrilateral.

Concept

A tangent is always perpendicular to the radius at the point of contact, so $\angle OPT = \angle OQT = 90°$ .
The four angles of quadrilateral OPTQ sum to 360°.

Key points

$90° + 90° + 120° + \angle PTQ = 360° \implies \angle PTQ = 60°$ — option (a).
This confirms the standard result $\angle PTQ + \angle POQ = 180°$ for two tangents from an external point.

Common mistakes

Assuming the angle between two tangents is always 90°, regardless of ∠POQ.
Arithmetic slips while summing the known angles (300° instead of the correct total), leading to distractors like 70° or 80°.
Forgetting the tangent–radius perpendicularity fact, without which the quadrilateral angle-sum method cannot be set up.

Common Mistakes

1Forgetting that the radius is perpendicular to the tangent at the point of contact.
2Assuming ∠PTQ is always 90° regardless of ∠POQ.
3Arithmetic errors while applying the quadrilateral angle sum property (360°) or the supplementary relation $\angle PTQ + \angle POQ = 180°$ .

Interesting Facts

The property that tangent ⊥ radius at the point of contact is a cornerstone result proved in Euclid's Elements (Book III) over two thousand years ago and remains central to CBSE's Circles chapter.

This exact result — $\angle PTQ + \angle POQ = 180°$ — is one of the most repeated 1-mark CBSE MCQs on the Circles chapter, appearing with different angle values across multiple board years.

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Frequently Asked Questions

What is the relationship between ∠PTQ and ∠POQ for two tangents from an external point?

$\angle PTQ + \angle POQ = 180°$ , because OPTQ is a quadrilateral with two right angles at P and Q (where the radius meets the tangent).

Why is the radius perpendicular to the tangent?

This is a fundamental circle theorem: at the point of contact, the tangent line is perpendicular to the radius drawn to that point, proved using the fact that the radius is the shortest distance from the centre to the tangent line.